Hello friends. So, welcome to a lecture series
on multivariable calculus. So, we were dealing with maxima minima of 2 or more than 2 variable
functions, now I will deal with Lagrange multipliers. So, what Lagrange multiplier method is let
us see. In many practical problems we need to find the maximum or minimum value of a
function. F X 1, X 2 up to X n when the variables are
not independent, but they are connected by one or more constraints of the form gi X 1,
X 2 up to Xn equal to 0 where i from 1 to k and generally n is greater than k.
Now, suppose you have a constraint optimization problem, we call such problems as constrained
optimization problem that is; you have to find out maximum and minimum value of some
function f subject to some conditions some conditions are given to you. Conditions are
gi of X 1, X 2 up to Xn equal to 0 ok. So, hence these variables X 1, X 2 up to Xn are
not independent they are connected by some relations. So, if they are connected by some
relations. So, how can you find out maximum or minimum value of function subject to those
conditions? How can we solve those problems? Suppose you have to find a point P xyz on
the plane x plus 2 y minus 3 z equal to 5 that is closest to the origin. So, what is the plane? Plane is x plus 2,
y minus 3, z equals to 5. Now in this plane there are so many points ok. You see this
plane this is the equation of plane is x plus 2, y minus 3, z equals to 5. From all those
points we have to find out that point which is closest or nearest from the origin ok.
So, this is this is some origin. So, let us suppose that point is x, y, z. So, distance
from the origin will be under root x minus 0 whole square, plus y minus 0 whole square
plus z minus 0 whole square. So, you have to you have to find out the minimum value
of this d subject to this condition ok. So, what is our problem? Problem is minimizing
this objective under root x square plus y square plus z square subject to x plus 2,
y minus 3, z equals to 5 or this problem can be rewritten as minimum of. Now finding minimum
of under root or finding minimum of the inside expression I wanted the same thing the problems
are equivalent. So, we can say that finding minimum of x square
plus y square plus z square subject to x plus 2, y minus 3, z is equal to 5. Now the first
way out is the method of substitution. So, what is a method of substitution? Now this
is an equation having 3 unknowns, you simply eliminate one other variable in this expression
in this expression with the help of this expression. So, let us compute say x. So, what will be x? It is 5 minus 2, y plus
3 z; so, what will be f? F will be 5 minus 2 y plus 3 z whole square plus y square plus
z square. So, now, it is something like unconstrained problem, I mean now it is a 2 variable problem
without any condition and you have to simply find a minimum value of this. So, the our
old technique will work out that is you find out the critical point of this function ok.
Another using the second order partial derivative test you can check where that point is a point
of local minima or local maxima. So, that will give the that will give the minimum value
of this function. So, what is fy and fz put it equal to 0. So,
this implies it is simply 2 times 5 minus 2 y plus 3 z into minus 2 plus 2 y is equal
to 0 and the second equation is 2 times 5 minus 2 y plus 3 z into 3 plus 2 z equal to
0. So, the equations are simply. So, equations are it is you can easily find out the 2 equation,
one equation is this second equation is this and when you solve these 2 equations. So,
you can find out the values of y and z of course, the second or derivative test will
give the minimum value I meanpoint as a local minima substituting those y and z over here
you can simply find out x. So, that xyz will be a point; which is closest
from the origin lying on the plane. So, what is that point? That point is simply 5 by 14, 5 by 7 minus
15 by 14 that you can easily find out solving these 2 equations simultaneously and then
find the values of y and z you simply substitute it over here find the value of x ok.
Now, to solve a constraint maximum or minimum problem by substitution do not always work
smoothly you see. Here we have a linear equation. So, we have easily substituted we have easily
find out one variable respect to other 2 and simply substituted over here, but this may
not always work. We have some non-linear terms also which is difficult to remove from this
expression in this expression you this relation ok. So, we have some method to find to solve
all those cases. So, that method is Lagrange multiplier method. Now what is that method? Now, suppose we want to maximize or minimize
the function fxy, we are considering here 2 variable function the same process will
go for 3 or more variable functions also subject to g x equal to 0 g x, y equal to 0. Now let
the extreme point of f x, y subject to the constraint g x, y equal to 0 is k and is attained
at x naught y naught. By examining the contours of f; we can see that at the extreme point
the curve g x, y equal to 0 must touch the level curve f x, y equal to k because if the
curve g x, y equal to 0 cut cross the level curve then one can a still move the point
along g x, y equal to 0. So, as to increase the decrease the value of f.
So, what does it mean? Let us see from the graph you see. Say this is this is the equation of g x, y
equal to 0. This graph which is given here in the bold line say this graph is a graph
of g x, y equal to 0 ok. Now you plot different contours of f say f x, y equals to m 1, f
x, y equals to m 2 f x, y equal to m 3 and so on ok. We can simply observe that at the
point x naught y naught if you draw a contour, which touches at this point will give the
stream values. Because if we if we if we go inside the region we are it cuts we are it
intersect g x, y equal to 0 then the value of the function may increase may increase
or decrease. I mean if you move further the value of value of the function may increase
the or decrease further. Suppose this for the point of local maxima,
where the function attained maximum value then if you go if you go in the region below
the curve value of the function will decrease and continuously decrease. So, this can be
understand by following say the first example let us discuss this example. So, what I want
to say basically, you see in the first example we have to find out the maximum minimum value
of the function and subject to condition is x square plus
y square equal to 1 ok. So, now this is our g; now let us draw different
contours of f. So, if you have. So, this is a straight line basically. So, add here, here
f equal to 3, x plus 4 y is 0; because it is passing through origin ok. Now when which
is 1; so suppose at this point it may be 12 when this is 4 and this is 3 at some point
here it may be 11, at some point here when it touches the circle it may have some other
value say k. Now when you when you come inside this circle
at this point and this point, the function have some value function 3, x plus 4 y have
some value. But as you move inside the function the value of the function decreases continuously
and it is 0 here. So, function will attain this function will attain maximum value when
it touches this circle over here. And again when you move inside this circle,
this will give the minimum value over here where it touches the when it touches the circle
over at this point ok. Because further decrement in the value of the function subject to this
condition is not possible, because if you take a function over here take a function
over here we want we want all those points which lies on the circle subject to this condition
we are we are talking about. We are talking about maximum minimum of this f over this
condition; I mean all those points lying on this circle.
Now, if we are saying that this is not a point of maxima this is the point of maxima. So,
all those values when we move inside the region, the value of f decreases here it is 0 and
when you away from this point, value of f increases ok, but we are interested to find
out those values those x and y, which are on the circle y square equal to 1. So, now,
at this point if you are talking about this point at this point, this will be simply gradient
of f; gradient of f is always normal to a surface normal to a curve.
And at this point at this point the this is also the gradient of g, the same direction
with the rate of g, g is g here is x square plus y square minus 1 equal to 0. Now at this
point gradient of g and gradient of f are same, I mean direction are same this means
they are parallel. And parallel means parallel means gradient of f will be some lambda times
gradient of g similarly here, gradient of f and gradient g are parallel.
So, this means at the point of at the point of extrema, if you have to maximize or minimize
the function subject to some condition g equal to 0; then at that point gradient of f is
always equal to lambda times gradient of g, because gradient of f and gradient of g are
parallel vectors. So, they are parallel this means one vector can be written as lambda
times other vector ok. So, this is the main concept of Lagrange multiplier method basically.
Now, here what is gradient of f? It is 3 i cap plus 4 j cap what is gradient of g? It
is 2 xi cap plus 2 yj cap del g by del x i cap plus del g by del y k cap and lambda is
called Lagrange multiplier this lambda is called Lagrange multiplier. So, what is gradient
of f is equal to lambda tangent of g. So, this implies 3 i cap plus 4 j cap will be
equals to 2 xi cap plus 2 yj cap this implies lambda times this implies 2 x lambda will
be equals to 3 and 2 y lambda will be equals to 4. So, here x will be equals to 3 upon
2 lambda and here y will be equals to 2 upon lambda.
Now, this point must be on circle also. So, this must satisfy x square plus y square equal
to 1 also. So, when you take x square plus y square equal to 1. So, this implies 9 by
4 lambda square plus 4 by lambda square will be equals to 1. This means 9 by 16, 9 plus
16 will be equals to 4 lambda square and this implies lambda square equals to 25 by 4 implies
lambda is equals to plus minus 5 by 2 ok. Now, a lambda is 5 by 2. So, we were solving
this problem that is maximizing and minimizing f x, y equal to 3 x plus 4 y subject to x
square plus y square equal to 1. And we have just obtain that x equals to 3 by 2 lambda
and y equal to 2 by lambda where lambda is plus minus 5 by 2. So, what your values of
x and y? X will be when you substitute. So, for lambda equals to 5 by 2 x will be equal
to it is 3 by 5 and y will be equal to 4 by 5 and for lambda equal to minus 5 by 2 x will
be minus 3 by 5 and y will be minus 4 by 5 ok.
So, where these points are? These points are on the circle, they say a unit circle of center
0 radius 1 it is 1 comma 0, minus 1 comma 0 it is 0 comma 1 and 0 comma minus 1, and
this point is somewhere here which is 3 by 5 and 4 by 5 and this is a point where this
function attains where this function attains maximum value what is that value? When you
substitute x as 3 by 5 and y as 4 by 5 it will be nothing, but 9 plus 16, 25 by 5 it
is 5. So, that is the maximum value of f and what
is the minimum value of f? Minimum value is obtained at x equals to minus 3 by 5 and y
equals to minus 4 by 5 and where this point is? This point is somewhat here which is minus
3 by 5 and minus 4 by 5. So, the this is this is the f which is a minimum value of this
function. So, the maximum value of function is 5, which is at this point and the minimum
value of function is minus 5 which is at this point.
So, let us try to solve a problem 2; the temperature at a point xy on a metal plate is given by
T x, y is equal to 4 x square minus 4 x, y plus y square. An ant on the plate walks around the circle
of radius 5 centered at the origin. So, we have to find the highest and the lowest temperatures
encountered by the ant. So, you have a metallic plate, the temperature of the metallic plate
is governed by that expression and ant is moving on that plate following a circular
path, circular path is centered it origin and radius is 5. So, basically we have to find out the maximum
and minimum value of T x, y which is given as 4 x square minus 4 xy plus y square subject
to x square plus y square is equal to 25 because ant is ant is following this root, ant is
following this circle is moving on this circle and when ant is feeling the maximum temperature
and when it is feeling at the lowest temperature, the points that we have to find out. So basically
we have to maximize or minimize this function subject to this condition.
So, how can we do that? We can take gradient of T is equal to lambda times gradient of
g; where g is x square plus y square minus 25 is equal to 0 this is by Lagrange multiplier
method what is gradient of T? It is 8 x minus 4 y i cap plus minus 4 x plus 2 y j cap is
equals to lambda times 2 xi cap plus 2 y j cap. So, this implies 8 x minus 4 y it equals
to 2 lambda x and minus 4 y plus 2 y is equals to lambda into 2 y.
So, from these 2 equation this implies x lambda by 2 will be equals to 2 x minus y from here
and from here it is minus 2 x plus y which is equals to lambda y. So, from these 2 equation
what we have obtained this implies x lambda by 2 is equals to minus y lambda and this
implies lambda times x plus 2 y is equal to 0.
So, now we have two possibilities; either lambda equal to 0 or x plus 2 y equal to 0.
So, first we will take lambda equal to 0. So, case one when lambda equal to 0. If lambda
equal to 0 this means 2 x minus y equal to 0, the same equation obtained from this expression.
So, 2 x minus y equal to 0 this implies. So, y is equals to 2 x now this expression this
x and y must satisfy this equation also because ant is moving on the circle. So, we substitute
y equal to 2 x here. So, it will be x square plus 4 x square is equal to 25. So, 5 x square
is equal to 25, and this implies x is equal to under plus minus under root 5.
So, what is the point? Point will be when x is under root 5 y will be 2 under root 5
and when x is minus under root 5, y will be minus 2 under root 5 and what will be T at
this point? T at this point will be when you substitute. So, so what is T? T is basically
when you carefully see on T; it is 2 x minus y whole square and 2 x minus y is 0. So, at
both this point T will be 0. So, this means this point and this point gives the minimum
value of the temperature or temperature is lowest at these points ok.
Now, now the case 2 when x plus 2 y equal to 0. If x plus 2 y equal to 0 means x is
minus 2 y. Now when you substitute x equal to x equal to minus 2 y here, it is 4 y square
plus y square equal to 25. So, this implies y is equal to plus minus under root 5. So,
therefore, then the points will be when y is plus under root 5. So, x will be minus
2 under root 5 and when is when it is minus under root 5, then it is 2 under root 5 and
values of T at this point will be when you substitute x as minus 2 under root 5 and y
as under root 5. So, this value will be this value will be 2 into 2 4 under root 5 plus
5 that is 125 and here also it is 125. So, this is the point these are the points, where
T is maximum. So, when ant is moving on the circle, at when
the ant comes at this point or this point the temperature is minimum that is T equal
to 0. And when the ant comes at this point or this point then temperature is maximum
which is 125 ok. So, that is how we can solve this problem. So, now, next problem find the
volume of the largest closed rectangular box in the first octant having 3 faces in the
coordinate planes and the vertex on the plane this ok.
So, have you have to find out the volume of the largest rectangular box can satisfy this
condition. So, volume of a rectangular box will be given
by x into y into z length into breadth into height and a subject to what is the condition.
So, this we have to maximize yes. So, this we have to maximize and the condition is basically
x by 1 plus y by 2, plus z by 3 is equals to 1 a subject to this condition we have to
maximize this value. So, again we will take. So, let us suppose
this is g. So, gradient of V will be equals to lambda times gradient of g and this implies
yz i cap plus zx j cap plus xy k cap will be equal to i cap plus 1 by 2 j cap plus 1
by 3 k cap lambda times. So, yz equal to lambda xz equals to lambda by 2 and xy is equals
to lambda by 3. Now fear from here what we obtain? You multiplied this by x this by y
this by z ok. So, what you will obtain? Xyz will be equals to x lambda will be equals
to y lambda by 2 will be equals to z lambda by 3.
Now, this point this condition must satisfy this also. So, what we are having from here?
It is x by 1 plus y by 2 plus z by 3 is equals to 1 you multiply the entire expression by
lambda. So, it is x by lambda into 1 by 1, y lambda by 2 plus z lambda by 3 is equal
to lambda say this is say this equals to k. So, this is a k plus k plus k is equal to
lambda. So, this implies lambda equals to 3 k ok.
So, from here what we obtain? X lambda is equals to y lambda by 2 is equals to z lambda
by 3 will be equals to 3 will be equals to lambda by 3. So, what will be x now? So, x will be x will be equals to 1 by 3,
y will be equals to 2 by 3 and z will be equals to 1 ok. So, you can simply substitute you
can simply check here. So, x 1 by 3 plus 1 by 3 plus 1 by 3 is one 1 minus 1 is 0. So,
satisfy this equation also ok. So, this is the point where it attains maximum
value and the maximum value of v will be nothing, but 1 by 3 into 2 by 3 into 1 that is 2 by
9. So, that is how we can find out we can solve this problem. Now suppose you have 2
constraints, you have 2 maximize maximum of you have to find out the maximum minimum value
of function fx yz subject 2 conditions. So, how can you find how can we use Lagrange multiplier
in this case? We can simply use we can simply take here
gradient of f as linear combinations of gradient of gradient of g 1 plus gradient of g and
gradient of g 2; that is lambda time gradient g 1 plus mu times gradient g 2 so, and this
condition also must hold that is we take a gradient of f as a linear combination of gradient
of the constraints gradient of g 1, gradient g 2, gradient g 3 and so on.
Now, let us solve the first problem in this case. So, here we have to find out the minimum
value of this function. Function is x square plus y square plus z
square and subject to condition are conditions are x plus 2 y plus 3 z equals to 6 and second
condition is x plus 3 y plus 9 z is equal to a 9. So, how can we solve this? So, gradient
of f we can take as lambda time gradient of g 1 plus mu time gradient of g. So, what is
g 1 from here? G 1 will be x plus 2 y plus 3 z minus 6 equal to 0 and g 2 from here will
be x plus 3 y plus 9 z minus 9 equal to 0. So, that will be gradient of g will be 2 x
i cap plus 2 y j cap plus 2 z k cap is equals to lambda times i cap plus 2 j cap plus 3
k cap plus mu times i cap plus 3 j cap plus 9 k cap. So, you formulate from the equations
from here it is 2 x equals to lambda plus mu, 2 y is equals to lambda 2 lambda plus
3 mu, and 2 z equal to 3 lambda plus 9 mu and next is these 2 condition also must hold.
So, you simply substitute x here y here and z here again here also you some simply substitute
the values of x y and z. So, you will get back 2 equations in lambda and mu you solve
those 2 questions find out the values of lambda and mu. When you get the values lambda and
mu is simply substitute here you find the values of xy and z that will give the point
where this f attained the minimum value. So, this is how we can solve this type of problems
a similarly we can solve the second problem also so.
Thank you.
